PDF of the exponential of the sum of $N$ independent gaussian random variables Is it true that the PDF of the sum of independent gaussian and identical random variables is gaussian with mean $N\cdot\mu$ and variance $N \cdot \sigma^2$? And if so how do we get the PDF of the exponential of such a sum?
 A: Let $x$and $y$ be Gaussian random variables; $x \sim \mathcal{N}(x; \mu_x, \sigma_x^2)$ and $y \sim \mathcal{N}(y; \mu_y, \sigma_y^2)$.
We know the Gaussian random variable's PDF is an exponential function. For instance:
$$
p(x) = \frac{e^{-\frac{1}{2\sigma_x^2}(x-\mu_x)^2}}{\sqrt{2\pi} \sigma_x},
$$
and similarly for $p(y)$.
Then, consider the sum $z = x + y$.
PDF of $z$ is a joint PDF of $x$ and $y$ and by independence,
$p(z) = p(x,y) = p(x)\cdot p(y)$.
We see then $p(z)$ will be a multiplication of two exponentials $p(x)$ and $p(y)$ and you know that exponents will add up to form a new exponential function.
By performing the completion of squares, you will get a new PDF that looks just like a Gaussian PDF with mean $\mu_z$ and the variance $\sigma_z^2$.
Now, for independent gaussian, these can be computed as follows:
$$
\mu_z = \mathbb{E}[z] = \mathbb{E}[x + y] = \mathbb{E}[x] + \mathbb{E}[y] = \mu_x + \mu_y,
$$
$$
\sigma_z^2 = var[z] = var[x + y] = var[x] + var[y] = \sigma_x^2 + \sigma_y^2.
$$
Assume now that $x$ and $y$ are actually identical random variables. Then, the above will be $N\mu$ and $N\sigma^2$ for $N$ identical independent Gaussian random variables.
Then, you can write now the PDF of $z$ just like how you would write the PDF for any Gaussian random variable:
$$
p(z) = \frac{e^{-\frac{1}{2\sigma_z^2}(z-\mu_z)^2}}{\sqrt{2\pi} \sigma_z},
$$
or more explicitly to your question, let $X$ be the sum of independent identical Gaussian random variables $x_i$ for $i = 1$ to $i = N$ so that  $X = \sum_{i=1}^{N}{x_i}$ and each $x_i$ having the mean and the variance as $\mu_x$ and $\sigma_x$.
$$
p(X) = \frac{e^{-\frac{1}{2N\sigma_x^2}(X-N\mu_x)^2}}{\sqrt{2\pi N} \sigma_x},
$$
A: Let $S=\sum X_k$.  The cdf for $e^S$ can be expressed as $F(x)=P(e^S\le x)=P(S\le ln(x))=\frac{1}{\sqrt{2\pi N}\sigma}\int\limits_{-\infty}^{ln(x)}e^{-\frac{(u-N\mu)^2}{2N\sigma^2}}du$
Thus the pdf $f(x)=\frac{1}{\sqrt{2\pi N}x\sigma}e^{-\frac{(ln(x)-N\mu)^2}{2N\sigma^2}}$
